Calculation of force due to electric dipole

  • #1
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Homework Statement


Two identical co-axial rings ,(radius R each) are kept separated by a small distance d, one of them carrying a charge +Q and the other a charge -Q. The charges are uniformly distributed over the respective rings. A point charge q is kept on the common axis of the rings, at a distance R from midpoint of their centers O. The net force on the charge q is (d<<R) (See image of solution)


Homework Equations


Electric field due to dipole on the axis = 2kp/(r^3) (r>>x)

The Attempt at a Solution


I have been taught that if charge and mass density in a body are distributed in same way then I can use position of centre of mass as the position where body acts as if its all charge is concentrated at that position.

Using that result gives me to points where I can assume that system of co-axial rings act as a dipole and electric field due to it and hence force may be calculated at any point.

Seems easy enough to apply in given question where centre of mass is situated at centre of ring.

p= Qd , r=R , E=2kQd/R^3. Clearly incorrect! (Does not tally with given answer!)

So, here is my request -
1. Would someone please explain the answer in image (The part dealing with cos^2 and sin^2)
2. Where is my method incorrect.
 

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Answers and Replies

  • #2
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Do not use the dipole thing. Use the principle of super-position. The cos and sin are there probably as part of the derivation. I already know a result for the field on the axis of a ring, so I used it. But yes, it definitely involves those values.

PS: I did not get the same result. My answer is four times as large.
 
  • #3
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6
Do not use the dipole thing. Use the principle of super-position. The cos and sin are there probably as part of the derivation. I already know a result for the field on the axis of a ring, so I used it. But yes, it definitely involves those values.

PS: I did not get the same result. My answer is four times as large.
Did you by any chance use, E= KQx/(R^2 + x^2)^3/2 for both rings and substract (or add depending upon you, if you think in terms of magnitude or vectors) and use binomial expansion?

My point is - that is how we deduce electric field due dipole - so using superposition, or result of dipole should not really matter, or maybe I got something wrong?
 
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  • #4
BvU
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You are getting closer. The ring approach is considerably easier: You have an expression for the on-axis E field -- use the one that still has the ##z## (the horizontal coordinate in your picture) in it. You need to work out ##E^+ - E^-## for small ##d##. In other words ##E(z) - E(z+d)##. Doesn't that remind you of the expression for the derivative of ##E## :rolleyes: ?

(I don't think it's called binomial expansion...)

For the dipole approach you can see that all the little dipoles formed by the charges on the + ring plus their nearest counterpart on the - ring are 'seen' under an angle of 45 degrees. So you need the expression for the off-axis field, then take the z (horizontal) component and then can add up (integrate) around the ring. (The other components cancel). The integration is easy: ##\int dp = {\rm d} \int dq = {\rm d} \int \lambda R\, d\phi = {\rm d} \,2\pi\lambda R= {\rm d}\, Q##
 
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  • #5
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For the dipole approach you can see that all the little dipoles formed by the charges on the + ring plus their nearest counterpart on the - ring are 'seen' under an angle of 45 degrees.
So basically you mean that we can't take the whole rings themselves as dipoles, right?
 
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  • #6
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Indeed you can not. That would ruin the calculation
 
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  • #7
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Thanks a lot guys. Apparently I should add a word of caution while treating system of charges as dipoles, I got the answer by both electric field and using dipoles method. Winding up both 11th and 12th class isn't the easiest thing to do, I have been using binomial theorem to get approx. results in thermodynamics so often that it never occured to me that I should just get away by differentiating the expression, much better and easier approximation.
 

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